My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?
thank you!
Hello,
this is only a guess but...
If the vertex has the coordinates V(k, h) the the equation of the parabola is:
$\displaystyle y = a(x-k)^2 + h$ . Equation in turning point form.
To calculate the intercepts with the x-axis you use that y = 0:
$\displaystyle 0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}$
Maybe this helps a little bit.
Hello,
here comes your example:
$\displaystyle y = -\frac{1}{4} \cdot (x-3)^2 + 4$
The vertex is at V(3, 4) and according to the formula given in my previous post you'll get the zeros Z_1 and Z_2 at:
$\displaystyle x_{1,2} = 3 \pm \sqrt{\frac{-4}{-\frac{1}{4}}} = 3 \pm 4$
I've attached the graph of the parabola so you can check the answer.